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  "body": "\n## Proof\n\nWe identify the derivative of a [[curve|0G7WM8DX4P]] with a linear map on $\\mathbb{R}$, and we apply the chain rule.\n\n1. Let $I$, $U$, $\\gamma$, $f$ and $t_0$ be as in the statement. The interval $I$ is an open subset of $\\mathbb{R} = \\mathbb{R}^1$.\n2. Let $\\sigma \\colon I \\to \\mathbb{R}^p$ be any map, let $t \\in I$, and let $c \\in \\mathbb{R}^p$. For real $s \\ne 0$ with $t + s \\in I$,\n$$\\frac{\\|\\sigma(t + s) - \\sigma(t) - s c\\|}{|s|} = \\left\\| \\frac{\\sigma(t + s) - \\sigma(t)}{s} - c \\right\\|.$$\n3. By step 2, $\\sigma'(t)$ exists and equals $c$ if and only if $\\sigma$ is [[differentiable|0135XZZQ8M]] at $t$ with $D\\sigma(t)(s) = s c$ for $s \\in \\mathbb{R}$. The maps $s \\mapsto s c$ are exactly the linear maps $\\mathbb{R} \\to \\mathbb{R}^p$.\n4. By step 3 with $\\sigma = \\gamma$, the curve $\\gamma$ is differentiable at $t_0$ with $D\\gamma(t_0)(s) = s \\, \\gamma'(t_0)$.\n5. The map $\\gamma$ goes from the open set $I \\subseteq \\mathbb{R}$ into $U$, and $f$ is differentiable at $\\gamma(t_0)$. By the [[chain rule in several variables|0EAAS2EHDR]] with $n = 1$, $f \\circ \\gamma$ is differentiable at $t_0$.\n6. The chain rule also gives, for $s \\in \\mathbb{R}$, $D(f \\circ \\gamma)(t_0)(s) = Df(\\gamma(t_0))\\big(s \\, \\gamma'(t_0)\\big) = s \\, Df(\\gamma(t_0))\\big(\\gamma'(t_0)\\big)$. The last equality holds since $Df(\\gamma(t_0))$ is linear.\n7. By step 3 with $\\sigma = f \\circ \\gamma$ and $c = Df(\\gamma(t_0))(\\gamma'(t_0))$, the derivative $(f \\circ \\gamma)'(t_0)$ exists and equals $Df(\\gamma(t_0))\\big(\\gamma'(t_0)\\big)$.\n\n## Idea\n\nA curve has a velocity vector $\\gamma'(t_0)$, and this is the same as a linear map $s \\mapsto s \\gamma'(t_0)$ from $\\mathbb{R}$. With this dictionary, the result is the general chain rule with one input variable. Feeding the velocity into $Df$ gives the rate of change of $f$ along the curve.\n",
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